I am using Kreyszig's functional analysis it is example in it which says that $l_p : (p≠2)$ is not an inner product space. For proof it uses the standard norm on $l_p$ and shows that it does not satisfies parallogram identify.
My question is that what about if we take other norm for example supremum norm on $l_p$, in fact we does not know how many norms we can have on $l_p$ then how checking for only one norm will guarantees it!
Thanks in advance
A normed linear space comes with a (real or complex) vector space and a norm. An inner product space comes with a vector space and an inner product. We can understand inner product spaces as normed linear spaces, since inner products induce a norm $\|v\| = \sqrt{\langle v, v \rangle}$.
When we say a normed linear space $(X, \| \cdot \|)$ is an inner product space, we (very slightly) abuse terminology. What we mean is that the norm on our normed linear space is induced by some inner product. That is, there exists an inner product $\langle \cdot, \cdot \rangle$ on $X$ such that $\|x\| = \sqrt{\langle x, x\rangle}$ for all $x \in X$. This is true if and only if the norm $\| \cdot \|$ satisfies the parallellogram identity (the "only if" is expected, but the "if" is kinda cool!).
In this case, the $(\ell^p, \| \cdot \|_p)$ spaces have their own norm equipped, as all normed linear spaces do. If you consider the underlying vector space with a different norm (e.g. the supremum norm), then it is a different normed linear space, and may be an inner product space (it's not hard to see that it isn't for this particular example, but you don't know in general).
As an easy example of this, you can put the $2$-norm on any $\ell^p$ space for $1 \le p < 2$. These are not inner product spaces with their bundled norms, but if you put the $2$-norm on them, then they are inner product spaces.